Olympic College Topic 18 Complex Numbers
Topic 18 Complex Numbers.
Consider the following quadratic equation x2 + 1 = 0
If we try to solve it we get the following situation:x2 + 1 =
x2 =
x =
0
–1
Subtracting 1 from both sides
Taking the square root of both sides
There are no real number solutions to this problem since the square of all real numbers will
be a positive number.
Mathematicians looked at this situation and created a new number, a complex number ,whose
value is defined as being equal to
and is often denoted by the letter i. This new number
i also has the interesting property that its square i2 = – 1 . From this new number i we can
create a complete number system made up of all possible combinations of a + bi where a and
b are real numbers. These numbers are collectively called the complex numbers with the term
a called the real part and the term b is called the imaginary part. For example, 5 + 2i is a
complex number where 5 is called the real part and 2 is called the imaginary part.
1. How to Add and Subtract Complex Numbers.
Example 1: Express
Solution:
as a complex number a + bi.
To express
as a complex number we split it into two parts 16 and – 1
This allows us to write
=
=
=
Example 2: Express
Solution:
as a complex number a + bi.
To express
as a complex number we split it into two parts 20 and – 1
This allows us to write
The
=
=
=
can in its turn also be simplified into
=
=
=
If we now combine these two results we get
=
Page | 1
Olympic College Topic 18 Complex Numbers
Example 3: Write each of the following as a complex number in the form a + bi.
(a) 4 +
Solution (a):
Solution (b):
Solution (c):
Solution (d):
9
(b)
(a)
(d)
(e) (3 + 2i) + (4 + 5i)
(f) (3 + 2i) – (4 + 5i)
(g) 3(3 + 2i) + 2(4 – 3i)
(h) 4(3 + 5i) – 2(4 – 6i)
4+
4 +
4+
 36 +
9
 16
4
=
4+
=
=
4+
4 + 3i
9  1
9  1
4  1 + 16  1
=
=
=
=
4   1 + 16   1
2i + 4i
6i
=
2+
=
=
2+ 4i
2 + 2i
 25 =
=
=
4  (1)
36 i +
6i + 5i
11i
25 i
Solution (e): (3 + 2i) + (4 + 5i) =
=
3 + 4 + 2i + 5i
7 + 7i
Collect like terms
Add and subtract like terms
Solution (f): (3 + 2i) – (4 + 5i) =
=
=
3 + 2i – 4 – 5i
3 – 4 + 2i– 5i
– 1 – 3i
Subtracting 4 + 5i
Collect like terms
Add and subtract like terms
Solution (g): 3(3 + 2i) + 2(4 – 3i) =
=
=
9 + 6i + 8 – 6i
9 + 8 + 6i – 6i
17
Using the distributive law
Collect like terms
Add and subtract like terms
Solution (h): 4(3 + 5i) – 2(4 – 6i)
12 + 20i – 8 – 12i
12 – 8 + 20i – 12i
4 + 8i
Using the distributive law
Collect like terms
Add and subtract like terms
=
=
=
Page | 2
Olympic College Topic 18 Complex Numbers
2. How to multiply two complex numbers.
Example 4: Express the following multiplications as a complex number a + bi.
(a) 2i(3 + 4i)
(b) 5i(3 + 4i)
(c) (2 + i)(4 – 3i)
(b) (5 + 4i)(5 – 4i)
(e) (2 + 3i)2
(f)
Solution (a): 2i(3 + 4i)
Solution (b): 5i(3 + 4i)
=
=
=
=
6i + 8i2
6i + 8(–1)
6i – 8
– 8 + 6i
Using the distributive law
Using the property that i2 = – 1
=
=
=
=
15i + 20i2
15i + 20(–1)
15i – 20
– 20 + 15i
Using the distributive law
Using the property that i2 = – 1
Solution (c): (2 + i)(4 – 3i)
Rearranging terms
Rearranging terms
=
8 – 6i + 4i – 3i2
= 8 – 2i – 3(–1)
=
=
8 – 2i + 3
11 – 2i
=
25 – 20i + 20i – 16i2 Using FOIL
= 8 – 2(–1)
Using the property that i2 = –
=
=
8+2
10
=
=
=
=
=
(2 + 3i)(2 + 3i)
4 + 6i + 6i + 9i2
4 + 12i + 9(–1)
4 + 12i – 9
– 5 + 12i
Using FOIL
Using the property that i2 = –
1
Solution (d): (5 + 4i)(5 – 4i)
1
Solution (e):
Solution (f):
(2 + 3i)2
Using the distributive law
Using the property that i2 = – 1
Adding and subtracting like terms
First we convert all the numbers to be in terms of i
=
=
=
=
=
(
-
)
Using the distributive law
i
=
Using the property that i2 = – 1
– 4 – 12i
Page | 3
Olympic College Topic 18 Complex Numbers
3. Solving Quadratic equations with Complex Numbers as Roots.
Some quadratic equations have no real numbers as roots, but we can in fact solve these
equations if we allow complex numbers to be solutions. The method is the same as that used
in all other situations the only difference is that the solutions are complex numbers.
Example 5: The quadratic equation x2 – 4x + 6 = 0 has two solutions which are both
complex numbers. Use the quadratic formula to find these solutions.
Solution:
x2 – 4x + 6 = 0
a=1b=–4c=6
b2 – 4ac =
So
x
(–4)2 – 4(1)(6)
b 2  4ac =
=
8
=
=
16 – 24 =
=
8i
2 2i
 (4)  2 2 i
2(1)
 b  b 2  4ac
=
2a
–8
42 2 i
2
=
42 2 i
= 2– 2i
2
42 2 i
and
x =
= 2+ 2i
2
There are 2 complex number solutions they are x = 2 – 2 i and x = 2 + 2 i
The two solutions are x
=
Example 6: The quadratic equation x2 – 2x + 2 = 0 has two solutions which are both
complex numbers. Use the quadratic formula to find these solutions.
Solution:
x2 – 2x + 2 = 0
a=1 b=–2 c=2
b2 – 4ac =
So
x
(– 2)2 – 4 x 1 x (2)
b 2  4ac
=
=
4
 b  b 2  4ac
2a
=
=
=
4–8 =–4
4  1 =
4  1
=
2i
2  2i
2
2(1  i )
(1  i )
2  2i
=
=
= 1+i
2
1
2
2(1  i )
(1  i )
2  2i
or x =
=
=
= 1–i
1
2
2
There are two complex numbers as solutions they are x = 1 + i and x = 1 – i
So x
=
Page | 4
Olympic College Topic 18 Complex Numbers
Exercise 1
1.
Write each of the following as a complex number in the form a + bi.
(a) 2 +
4
(b)
(c) (2 + 5i) + (5 – 5i)
(d) (1 + 2i) – (3 + 4i)
(e) 3(4 + i) + 5(1 – 4i)
(f) 2(4 + 6i) – 2(3 – 2i)
2.
Express the following multiplications as a complex number a + bi.
(a) 5i(3 + 8i)
(b) –i(4 + 2i)
(c) (4 + 2i)(1 – 2i)
(d) (5 + i)( –5 – i)
(e) (3 – 2i)2
(f)
3.
The quadratic equation x2 + 36 = 0 has two solutions which are both complex numbers
Use the quadratic formula to find these solutions.
4.
The quadratic equation 2x2 – 4x + 8 = 0 has two solutions which are both complex
numbers
Use the quadratic formula to find these solutions.
5.
The quadratic equation 9x2 – 6x + 2 = 0 has two solutions which are both complex
numbers
Use the quadratic formula to find these solutions.
Page | 5
Olympic College Topic 18 Complex Numbers
Solutions:
Exercise 1: 1(a) 2 + 2i (b) 9i (c) 7 (d) – 2 – 2i (e) 17 – 16i
(f) 2 + 20i
2.(a) – 40 + 15i (b) 2 – 4i (c) 6 – 6i (d) – 24 – 10i
(e) 5 – 12i (f) – 18
3. x = 6i and x = – 6i
4. x = 2 + 2i and x = 2 – 2i
5. x =
and
x=
Page | 6